3.75.22 \(\int \frac {e^{\frac {12}{(-15+20 x) \log (x \log ^2(2))}} (72-96 x-96 x \log (x \log ^2(2)))}{(45 x-120 x^2+80 x^3) \log ^2(x \log ^2(2))} \, dx\)

Optimal. Leaf size=28 \[ -1+2 \left (-2+e^{\frac {3}{5 \left (-\frac {3}{4}+x\right ) \log \left (x \log ^2(2)\right )}}\right ) \]

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Rubi [F]  time = 2.52, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(12/((-15 + 20*x)*Log[x*Log[2]^2]))*(72 - 96*x - 96*x*Log[x*Log[2]^2]))/((45*x - 120*x^2 + 80*x^3)*Log[
x*Log[2]^2]^2),x]

[Out]

(8*Defer[Int][E^(12/((-15 + 20*x)*Log[x*Log[2]^2]))/(x*Log[x*Log[2]^2]^2), x])/5 - (32*Defer[Int][E^(12/((-15
+ 20*x)*Log[x*Log[2]^2]))/((-3 + 4*x)*Log[x*Log[2]^2]^2), x])/5 - (96*Defer[Int][E^(12/((-15 + 20*x)*Log[x*Log
[2]^2]))/((-3 + 4*x)^2*Log[x*Log[2]^2]), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{x \left (45-120 x+80 x^2\right ) \log ^2\left (x \log ^2(2)\right )} \, dx\\ &=\int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{5 x (-3+4 x)^2 \log ^2\left (x \log ^2(2)\right )} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{x (-3+4 x)^2 \log ^2\left (x \log ^2(2)\right )} \, dx\\ &=\frac {1}{5} \int \frac {24 e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (3-4 x-4 x \log \left (x \log ^2(2)\right )\right )}{(3-4 x)^2 x \log ^2\left (x \log ^2(2)\right )} \, dx\\ &=\frac {24}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (3-4 x-4 x \log \left (x \log ^2(2)\right )\right )}{(3-4 x)^2 x \log ^2\left (x \log ^2(2)\right )} \, dx\\ &=\frac {24}{5} \int \left (-\frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{x (-3+4 x) \log ^2\left (x \log ^2(2)\right )}-\frac {4 e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )}\right ) \, dx\\ &=-\left (\frac {24}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{x (-3+4 x) \log ^2\left (x \log ^2(2)\right )} \, dx\right )-\frac {96}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )} \, dx\\ &=-\left (\frac {24}{5} \int \left (-\frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{3 x \log ^2\left (x \log ^2(2)\right )}+\frac {4 e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{3 (-3+4 x) \log ^2\left (x \log ^2(2)\right )}\right ) \, dx\right )-\frac {96}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )} \, dx\\ &=\frac {8}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{x \log ^2\left (x \log ^2(2)\right )} \, dx-\frac {32}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x) \log ^2\left (x \log ^2(2)\right )} \, dx-\frac {96}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 0.86 \begin {gather*} 2 e^{\frac {12}{5 (-3+4 x) \log \left (x \log ^2(2)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(12/((-15 + 20*x)*Log[x*Log[2]^2]))*(72 - 96*x - 96*x*Log[x*Log[2]^2]))/((45*x - 120*x^2 + 80*x^3
)*Log[x*Log[2]^2]^2),x]

[Out]

2*E^(12/(5*(-3 + 4*x)*Log[x*Log[2]^2]))

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fricas [A]  time = 0.61, size = 21, normalized size = 0.75 \begin {gather*} 2 \, e^{\left (\frac {12}{5 \, {\left (4 \, x - 3\right )} \log \left (x \log \relax (2)^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-96*x*log(x*log(2)^2)-96*x+72)*exp(12/(20*x-15)/log(x*log(2)^2))/(80*x^3-120*x^2+45*x)/log(x*log(2)
^2)^2,x, algorithm="fricas")

[Out]

2*e^(12/5/((4*x - 3)*log(x*log(2)^2)))

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giac [A]  time = 0.32, size = 27, normalized size = 0.96 \begin {gather*} 2 \, e^{\left (\frac {12}{5 \, {\left (4 \, x \log \left (x \log \relax (2)^{2}\right ) - 3 \, \log \left (x \log \relax (2)^{2}\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-96*x*log(x*log(2)^2)-96*x+72)*exp(12/(20*x-15)/log(x*log(2)^2))/(80*x^3-120*x^2+45*x)/log(x*log(2)
^2)^2,x, algorithm="giac")

[Out]

2*e^(12/5/(4*x*log(x*log(2)^2) - 3*log(x*log(2)^2)))

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maple [A]  time = 0.02, size = 22, normalized size = 0.79

method	result	size
risch	\(2 \,{\mathrm e}^{\frac {12}{5 \left (4 x -3\right ) \ln \left (x \ln \relax (2)^{2}\right )}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-96*x*ln(x*ln(2)^2)-96*x+72)*exp(12/(20*x-15)/ln(x*ln(2)^2))/(80*x^3-120*x^2+45*x)/ln(x*ln(2)^2)^2,x,meth
od=_RETURNVERBOSE)

[Out]

2*exp(12/5/(4*x-3)/ln(x*ln(2)^2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-96*x*log(x*log(2)^2)-96*x+72)*exp(12/(20*x-15)/log(x*log(2)^2))/(80*x^3-120*x^2+45*x)/log(x*log(2)
^2)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 5.04, size = 28, normalized size = 1.00 \begin {gather*} 2\,{\mathrm {e}}^{-\frac {12}{30\,\ln \left (\ln \relax (2)\right )+15\,\ln \relax (x)-40\,x\,\ln \left (\ln \relax (2)\right )-20\,x\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(12/(log(x*log(2)^2)*(20*x - 15)))*(96*x + 96*x*log(x*log(2)^2) - 72))/(log(x*log(2)^2)^2*(45*x - 120
*x^2 + 80*x^3)),x)

[Out]

2*exp(-12/(30*log(log(2)) + 15*log(x) - 40*x*log(log(2)) - 20*x*log(x)))

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sympy [A]  time = 0.52, size = 17, normalized size = 0.61 \begin {gather*} 2 e^{\frac {12}{\left (20 x - 15\right ) \log {\left (x \log {\relax (2 )}^{2} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-96*x*ln(x*ln(2)**2)-96*x+72)*exp(12/(20*x-15)/ln(x*ln(2)**2))/(80*x**3-120*x**2+45*x)/ln(x*ln(2)**
2)**2,x)

[Out]

2*exp(12/((20*x - 15)*log(x*log(2)**2)))

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